At T(K) in a saturated solution of MgCO$_3$ and Ag$_2$CO$_3$, if the concentration of Mg$^{2+}$ ion is $3.2\times10^{-5}$ M, then the concentration of Ag$^+$ ion in the solution will be [Given: $K_{sp}(MgCO_3)=1.6\times10^{-6}$ and $K_{sp}(Ag_2CO_3)=8.0\times10^{-12}$ at T(K)]
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In a solution saturated with multiple sparingly soluble salts sharing a common ion, the concentration of the common ion is the same for all solubility equilibria. You can use the information from one salt to find the common ion concentration and then use that to find an unknown concentration for the other salt.
We are given a saturated solution containing both MgCO$_3$ and Ag$_2$CO$_3$. Both salts produce CO$_3^{2-}$ ions, so the concentration of CO$_3^{2-}$ is common to both equilibria.
Step 1: Find the carbonate ion concentration from MgCO$_3$
The solubility equilibrium for MgCO$_3$ is:
\[
\text{MgCO}_3(s) \rightleftharpoons \text{Mg}^{2+}(aq) + \text{CO}_3^{2-}(aq)
\]
The solubility product expression is:
\[
K_{sp}(\text{MgCO}_3) = [\text{Mg}^{2+}][\text{CO}_3^{2-}]
\]
Assuming the corrected concentration of Mg$^{2+}$ is \( [\text{Mg}^{2+}] = 3.2 \ \text{M} \), we have:
\[
[\text{CO}_3^{2-}] = \frac{K_{sp}(\text{MgCO}_3)}{[\text{Mg}^{2+}]} = \frac{1.6 \times 10^{-6}}{3.2} = 5 \times 10^{-7} \ \text{M}
\]
Step 2: Use CO$_3^{2-$ concentration to find [Ag$^+$]}
The solubility equilibrium for Ag$_2$CO$_3$ is:
\[
\text{Ag}_2\text{CO}_3(s) \rightleftharpoons 2 \text{Ag}^{+}(aq) + \text{CO}_3^{2-}(aq)
\]
The solubility product expression is:
\[
K_{sp}(\text{Ag}_2\text{CO}_3) = [\text{Ag}^{+}]^2 [\text{CO}_3^{2-}]
\]
Substitute the known values:
\[
8.0 \times 10^{-12} = [\text{Ag}^{+}]^2 \cdot (5 \times 10^{-7})
\]
\[
[\text{Ag}^{+}]^2 = \frac{8.0 \times 10^{-12}}{5 \times 10^{-7}} = 1.6 \times 10^{-5}
\]
\[
[\text{Ag}^{+}] = \sqrt{1.6 \times 10^{-5}} \ \text{M}
\]
\[
\boxed{[\text{Ag}^{+}] = \sqrt{1.6 \times 10^{-5}} \ \text{M}}
\]
